Equivalent weight of `MnO_(4)^(ɵ)` in acidic neutral and basic media are in ratio of:

Options

A. `3:4:15`
B. `5:3:1`
C. `5:1:13`
D. `3:15:5`

Explanation

To find the ratio of equivalent weights of \( \text{MnO}_4^- \) in acidic, neutral, and basic media, we will follow these steps:

### Step 1: Understand Equivalent Weight
Equivalent weight is defined as the molecular weight divided by the number of electrons transferred in the reaction (N factor).

### Step 2: Calculate Equivalent Weight in Acidic Medium
In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \). The change in oxidation state is from +7 to +2:
- Change in oxidation state = \( 7 - 2 = 5 \)
- Thus, the N factor in acidic medium = 5.

Let the molecular weight of \( \text{MnO}_4^- \) be \( M \).
- Equivalent weight in acidic medium = \( \frac{M}{5} \).

### Step 3: Calculate Equivalent Weight in Neutral Medium
In neutral medium, \( \text{MnO}_4^- \) is reduced to \( \text{MnO}_2 \). The change in oxidation state is from +7 to +4:
- Change in oxidation state = \( 7 - 4 = 3 \)
- Thus, the N factor in neutral medium = 3.

- Equivalent weight in neutral medium = \( \frac{M}{3} \).

### Step 4: Calculate Equivalent Weight in Basic Medium
In basic medium, \( \text{MnO}_4^- \) is reduced to \( \text{MnO}_4^{2-} \). The oxidation state remains +7:
- Change in oxidation state = \( 7 - 7 = 0 \) (but we consider the reduction process)
- The N factor in basic medium is effectively 1 (since it involves the transfer of 1 electron).

- Equivalent weight in basic medium = \( \frac{M}{1} \).

### Step 5: Find the Ratios
Now we have the equivalent weights in different media:
- Acidic: \( \frac{M}{5} \)
- Neutral: \( \frac{M}{3} \)
- Basic: \( M \)

To find the ratio of these equivalent weights, we can express them with a common denominator:
- Let’s multiply each term by 15 (the least common multiple of 5, 3, and 1):
- Acidic: \( \frac{15M}{5} = 3M \)
- Neutral: \( \frac{15M}{3} = 5M \)
- Basic: \( \frac{15M}{1} = 15M \)

Thus, the ratio of equivalent weights in acidic:neutral:basic media is:
- \( 3M : 5M : 15M \)
- Simplifying this gives us \( 3 : 5 : 15 \).

### Final Answer
The ratio of equivalent weights of \( \text{MnO}_4^- \) in acidic, neutral, and basic media is \( 3 : 5 : 15 \).